Friday, May 4, 2012

Problem 749: Triangle, Rhombus, Parallel, Harmonic Means, Metric Relations

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 749 details.

Online Geometry Problem 748: Quadrilateral, 90 Degrees, Diagonal, Perpendicular, Metric Relations.

2 comments:

  1. Since ABEF is a rhombus so AE is an angle bisector of angle BAC
    We have BE/EC=AB/AC= c/b
    BE/BC= c/(b+c)=x/b => x=b.c/(b+c)

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  2. ADEF is a rhombus.FE = x = DE
    x/b = DE/AC = BE/BC and
    x/c = FE/AB = CE/BC
    Add:
    x/b + x/c = (BE + CE)/BC = BC/BC = 1
    x(1/b + 1/c)= 1
    x = bc /(b+c)
    Also
    x = (1/2)[2/(1/b + 1/c)]:=(1/2)HM(b,c)

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